Let $u \in \mathscr{E}'(\mathbb{R}^n)$ be a distribution with compact support, and let $\psi \in C_c^\infty(\mathbb{R}^n)$. I would like to show that
$$u * \psi = v $$ as tempered distributions, where $$\langle u * \psi, \phi \rangle := \langle u, \psi * \tilde{\phi} \rangle, $$ with $\tilde{\phi}(x) := \phi(-x)$, and
$$v(x) := \langle u, y \mapsto \psi(x – y) \rangle. $$
Observe that we can define the pairing between $u * \psi$ and an arbitrary distribution $f \in \mathscr{S'}(\mathbb{R}^n)$ as $$\langle u * \psi, f \rangle := \langle u, \left( x \mapsto \langle f, \psi(\cdot – x) \rangle \right) \rangle. $$
Using this definion, $$v(x) = \langle u * \psi, \delta_x \rangle, $$ where $\delta_x$ is the delta distribution at $x$. I then tried to show that for $\phi \in \mathscr{S}(\mathbb{R}^n)$, $$\langle u * \psi, \phi \rangle = \langle v, \phi \rangle, $$ by taking a mollifier $(\chi_\varepsilon)_{\varepsilon > 0}$, showing that $$\langle u * \psi, \chi_\varepsilon *\phi \rangle = \langle v, \chi_\varepsilon * \phi \rangle, $$ and then letting $\varepsilon \to 0$, but I have not managed to prove this.
Best Answer
Have a look at Proposition $9.3$ in Folland's Real analysis: Modern techniques and their applications (2nd edition). The idea is to approximate $$\langle v, \phi \rangle $$ with Riemann sums.